1
$\begingroup$

Suppose I have a function $f(\mathbf{x}(t))$, where $\mathbf{x}(t)$ is a vector. Assuming the derivative of $f$ at $t$ exists, I think it can be written, using the chain rule, as

$$\frac{df}{dt} = \frac{df}{d\mathbf{x}^{\mathrm{T}}} \frac{d\mathbf{x}}{dt} = \nabla_{\mathbf{x}} f \frac{d\mathbf{x}}{dt}$$

where $\nabla_{\mathbf{x}} f = \frac{df}{d\mathbf{x}^{\mathrm{T}}}$ is the gradient of $f$ (I've abused the derivative notation a little here but I think this is commonly used in vector calculus so should be ok).

I want to derive this result "rigorously" using limits, but I'm getting stuck. Here's my attempt. Starting with the outer function, $f$ is differentiable at $\mathbf{x}(t)$ if

$$f(x(t + \Delta t)) = f(x(t)) + \frac{df}{d\mathbf{x}^{\mathrm{T}}} \Delta \mathbf{x}(t) + o(||\Delta \mathbf{x}(t)||), \tag{1}\label{eq:1}$$

where $o(||\Delta \mathbf{x}(t)||)$ is a function such that $\lim_{\Delta \mathbf{x} \rightarrow \mathbf{0}} o(||\Delta \mathbf{x}(t)||) / ||\Delta \mathbf{x}(t)|| = 0$ and $\Delta \mathbf{x}(t) = \mathbf{x}(t + \Delta t) - \mathbf{x}(t)$. Now moving on to the inner function, $\mathbf{x}$ is differentiable at $t$ if

$$x(t + \Delta t) = x(t) + \frac{d\mathbf{x}}{dt} \Delta t + o(\Delta t),\tag{2}\label{eq:2}$$

where $o(\Delta t)$ is a function such that $\lim_{\Delta t \rightarrow 0} o(\Delta t) / \Delta t = 0$. This definition means I can also write

$$\Delta \mathbf{x}(t) = \frac{d\mathbf{x}}{dt} \Delta t + o(\Delta t). \tag{3}\label{eq:3}$$

Substituting this into the second term of \eqref{eq:1} I obtain

$$f(x(t + \Delta t)) = f(x(t)) + \frac{df}{d\mathbf{x}^{\mathrm{T}}} \frac{d\mathbf{x}}{dt} \Delta t + \frac{df}{d\mathbf{x}^{\mathrm{T}}} o(\Delta t) + o(||\Delta \mathbf{x}(t)||)\tag{4}\label{eq:4}.$$

If those last two terms simplify to $o(\Delta t)$ (which I'm under the impression they should - e.g. from this post) then I can obtain a limit definition for the derivative of $f(\mathbf{x}(t))$. I can't see how they simplify though. Is it just that they will disappear in the limit as $\Delta t \rightarrow 0$? It's easy to see that $\frac{df}{d\mathbf{x}^{\mathrm{T}}} o(\Delta t) / \Delta t$ will go to zero, but not obvious to me that $o(||\Delta \mathbf{x}(t)||) / \Delta t$ will also go to zero.

I'd love some help and also appreciate feedback on whether my derivation otherwise makes sense and is appropriately written. I'm not a mathematician so these kinds of formal treatment of problems is a bit alien to me. Any tips much appreciated. Thanks.

$\endgroup$
1
  • 1
    $\begingroup$ By the way, you're using $O$ (big-oh) where everyone else uses $o$ (little-oh). $\endgroup$ Commented Oct 10, 2017 at 23:41

1 Answer 1

2
$\begingroup$

Because $\mathbf x$ is differentiable, we know that $\dfrac{\|\Delta\mathbf x\|}{|\Delta t|}$ is bounded for $|\Delta t|$ small. Therefore, if $\text{blah} = o(\|\Delta\mathbf x\|)$, then $$\frac{\text{blah}}{\Delta t} = \frac{\text{blah}}{\|\Delta\mathbf x\|}\frac{\|\Delta\mathbf x\|}{\Delta t}.$$ The first term in the product goes to $0$ and the second term is bounded, so the product goes to $0$.

If it bothers you that you might be dividing by $0$, then write it without denominators. For small $\epsilon>0$, $$|\text{blah}|\le \epsilon \|\Delta\mathbf x\| \le \epsilon\left(\|\frac{d\mathbf x}{dt}\Delta t\| + |o(\Delta t)|\right) \le \big(\epsilon\|\frac{d\mathbf x}{dt}\| + \epsilon'\big)|\Delta t|,$$ and the stuff in parentheses can be made as small as you need by choosing $|\Delta t|$ small enough.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .